Physic Labs

Problem 3

On an optical bench, a thin converging lens of focal length f=10.0 cmf=10.0\,\mathrm{cm} is used with an object of height ho=2.00 cmh_o=2.00\,\mathrm{cm} placed on the principal axis, initially at distance u=30.0 cmu=30.0\,\mathrm{cm}. Work in air and in the paraxial approximation. For a real object take u>0u>0; a real image has v>0v>0. Define transverse magnification by M=hi/ho=−v/uM=h_i/h_o=-v/u. (a) Find the image distance and magnification, and state whether the image is real or virtual and upright or inverted. (b) Find the image height. (c) Move the object to u=15.0 cmu=15.0\,\mathrm{cm} and repeat the position, magnification, and image-character determination, then compare the two placements.
v2=(110.0−115.0)−1=30.0 cmv_2=\left(\frac1{10.0}-\frac1{15.0}\right)^{-1}=30.0\,\mathrm{cm}
problems.proof.analysis

At the new position u2=15.0 cm=1.5fu_2=15.0\,\mathrm{cm}=1.5f, the lens equation gives v2=30.0 cmv_2=30.0\,\mathrm{cm}. The image recedes as the object approaches the focal point from outside, consistent with v\toinftyv\toinfty as u→f+u\to f^+.